Maths problem: fatigue vs critical hits

[lch]

lch said:
Now with two dice, I'm not sure but I think it goes like this: You want a combination of two rolls that add up to X, so X = Y+(X-Y) and 0 < Y < X. Your probability is given as follows: For every Y from 1 to X-1, multiply P(Y) and P(X-Y) as given above. Add all of them together, divide by X-1. That's your probability for two rolls. You can simplify it into a closed form, which could be kinda tricky because of the modulo representation for Y and X-Y. Maple helps.

Okay, the second part was pretty bogus. It shows that I never really learned probability calculus. Another try:

X = 6a + b, 0 <= b < 6. Then the probability that a DRN gives X or higher on a roll is P(">= X") = (7-b)(1/6)^(a+1).
The probability that it equals X is P("= X") = (1/6)^(a+1).

But measuring a two-die roll isn't as easy. Here's a small python program that I used to check cleveland's calculations:

Code:

---

#!/usr/bin/env python



def p_1(x):

	if x < 1:

		return 1

	b = x % 6

	a = (x-b)/6

	return (7-b)*pow((float(1)/6), a+1)



def p_2(x):

	if x < 1:

		return 0

	b = x % 6

	a = (x-b)/6

	return pow((float(1)/6), a+1)



def doubledrn(x):

	M = [(a,b) for a in range(1,x-1) for b in range(1,x-a)]

	s = 0.0

	for (x_1, x_2) in M:

		s += p_2(x_1)*p_2(x_2)

	return 1-s



def doubledrn_2(x):

	s = p_1(x)

	for y in range(1,x-1):

		s += p_2(y)*p_1(x-y)

	return s



for x in range(2,18):

	print ">= %2d : %f" % (x, doubledrn(x))

---

And here's the output of it: Code:

---

>=  2 : 1.000000

>=  3 : 0.972222

>=  4 : 0.916667

>=  5 : 0.833333

>=  6 : 0.722222

>=  7 : 0.583333

>=  8 : 0.462963

>=  9 : 0.361111

>= 10 : 0.277778

>= 11 : 0.212963

>= 12 : 0.166667

>= 13 : 0.119599

>= 14 : 0.079475

>= 15 : 0.046296

>= 16 : 0.020062

>= 17 : 0.000772

---

So as far as I can tell, cleveland's table is right. The difference in later numbers probably comes from rounding precision being a ***** again, as my code probably isn't numerically stable. If I'd calculate greater values, the numbers even turn negative! I'd be interested what formulas cleveland used to be able to trick the rounding precision.

[cleveland]

lch said:
If I'd calculate greater values, the numbers even turn negative!

Negative probabilities? Sounds like my odds the last time I was in Vegas.

Here's roughly what I did:
1) Constructed a vector A(i) of the explicit probabilities for each single drn outcome i. This is pretty straightforward:
A(1)=A(2)=...=A(5)=1/6
A(6)=A(7)=...=A(10)=(1/6)^2
Etc.

2) Constructed a vector B(j) of the probabilities a second drn would be less than or equal to j. So:
B(1)=A(1)
B(2)=A(1)+A(2)
B(3)=A(1)+A(2)+A(3)
Etc.

3) Constructed a matrix M such that M(i,j) = A(i)*B(j). Therefore, each M(i,j) = P(X<=i+j | drn1=i).

4) Summing the / diagonals of this matrix (i.e. all M(i,j) whose i+j are identical) gives the probability that the 2d6oe will be less than or equal to i+j.

So for example, lets take the case when we want the 2d6oe <= 3. This is only possible 2 ways: if drn1=1 & drn2<=2, or if drn1=2 & drn2=1.

We therefore have:
A(1)=1/6
A(2)=1/6

B(1)=1/6
B(2)=2/6

M(1,2)=(1/6)*(1/6)
M(2,1)=(1/6)*(2/6)

So M(1,2)+M(2,1)= 0.083

Of course, this summation becomes larger for larger X=i+j.
When X=7, you'll need to add M(1,6)+M(2,5)+M(3,4)+...+M(6,1).

Phew.

[DonCorazon]

And now in English?

[cleveland]

DonCorazon said:
And now in English?

English? I didn't do very well in English...

[lch]

cleveland said:
Here's roughly what I did:
1) Constructed a vector A(i) of the explicit probabilities for each single drn outcome i. This is pretty straightforward:
A(1)=A(2)=...=A(5)=1/6
A(6)=A(7)=...=A(10)=(1/6)^2
Etc.

I think this is already where I fail, because the rest is pretty much the same. I was jumping back and forth between modulo 5 and modulo 6 just because the whole thing is offset by 1 and the gaps didn't work out. Let's correct this now...

Let X-1 = 5a + b, 0 <= b < 5. (notice the -1!!) Then the probability that a DRN gives exactly X is p(DRN = X) = (1/6)^(a+1).
The probability that it is X or greater is:


Now, we have two ways of measuring all the (x,y) in IN x IN which sum is equal to or greater than X: Either we take all the (x,y) with x+y < X and subtract the probability of them showing up simultaneously from 1, thus inverting the set, like you did if I understood right, or measure the following (with a loose mathematical notation): Code:

---

  (1, >= X-1)

  (2, >= X-2)

  (3, >= X-3)

   ... etc.

  (X-1, >= 1)

  (>= X, *)

---

which should be equal to {(x,y) | x+y >= X}. I implemented both versions in Python again: Code:

---

#!/usr/bin/env python



# p(DRN) >= x

def p_1(x):

	if x < 1:

		return 1

	b = (x-1) % 5

	a = (x-1) / 5

	return (6-b)*pow((float(1)/6), a+1)



# p(DRN) == x

def p_2(x):

	if x < 1:

		return 0

	b = (x-1) % 5

	a = (x-1) / 5

	return pow((float(1)/6), a+1)



# p(2d6oe) >= x

def doubledrn(x):

	M = [(a,b) for a in range(1,x-1) for b in range(1,x-a)]

	s = 0.0

	for (x_1, x_2) in M:

		s += p_2(x_1)*p_2(x_2)

	return 1-s



# p(2d6oe) >= x (alternate version)

def doubledrn_2(x):

	s = p_1(x)

	for y in range(1,x):

		s += p_2(y)*p_1(x-y)

	return s



for x in range(2,20):

	print ">= %2d : %f" % (x, doubledrn(x))

	print ">= %2d : %f" % (x, doubledrn_2(x))

---

and huzzah, they both give the same result, and it is: Code:

---

>=  2 : 1.000000

>=  3 : 0.972222

>=  4 : 0.916667

>=  5 : 0.833333

>=  6 : 0.722222

>=  7 : 0.583333

>=  8 : 0.462963

>=  9 : 0.361111

>= 10 : 0.277778

>= 11 : 0.212963

>= 12 : 0.166667

>= 13 : 0.127315

>= 14 : 0.094907

>= 15 : 0.069444

>= 16 : 0.050926

>= 17 : 0.039352

>= 18 : 0.029578

>= 19 : 0.021605

---

[lch]

lch said:
The probability that it is X or greater is: (...) (6-b)(1/6)^(a+1)

I just noticed that it is way easier to get the same result in a different way than I did. We already know that each element in a modulo "layer" shares the same probability, and that all the following layers together are cramped into something having the same probability, too. So you just need to count from the current element to the end and sum up.

[Pehmyt]

I don't know if you are still interested in this, but I was, as it says "maths problem" in the title For future reference, if nothing else, the actual probability distributions for 1d6oe and 2d6oe (=DRN in the manual) are:


and for the original question,

In all expressions X=5n+r, with 1<=r<=5 (note limits!).

The first one is more or less trivial, the second is proven by induction in n and the third follows from that by straightforward summation.

[Revolution]

I started reading all of this and thought of this picture...

[moderation]

So that is what kids are learning in school these days.